八 3. 連續型機率

八 _3. 連續型機率

3.1 (1)

因為f_X(t) = ^√1_πe^−t2 ≥ 0, t ∈ R,而且 R∞

−∞f_X(t)dt = R∞

−∞

√1

πe^−t2dt =R∞ 0

√1

πe^−t2dt +R0

−∞

√1

πe^−t2dt

= ^√1_π(R∞

0 e^−t2dt +R0

−∞e^−t2dt) = ^√1_π(R∞

0 e^−t2dt +R∞

0 e^−t2dt) = ^√1_π(

√π 2 +

√π 2 ) = 1 所以_fX(t)是機率密度函數

(2) fX(t) =







e^−t t ≥ 0 0 t < 0 因為_fX(t) ≥ 0, t ∈ R,而且 R∞

−∞f_X(t)dt = R∞

0 e^−tdt = lim

a→∞

Ra

0 e^−tdt = lim

a→∞(−e^−a+ 1) = 1 所以_fX(t)是機率密度函數

(3) f_X(t) =







√1

2πt⁻¹²e^−t2 t > 0

0 t ≤ 0

因為_fX(t) ≥ 0, t ∈ R,而且 R∞

−∞fX(t)dt = R∞ 0

√1

2πt⁻¹²e^−2tdt = ^√1_2πR∞

0 t⁻¹²e^−t2dt = ^√1_π R∞

0 u⁻¹²e^−udu = ^√1_π√ π

= 1 (u = ₂^t) (Γ(¹₂) = √

π) 所以_fX(t)是機率密度函數

(4) f_X(t) =







β^α

Γ(α)t^α−1e^−βt t > 0

0 t ≤ 0

因為_fX(t) ≥ 0, t ∈ R,而且 R∞

−∞f_X(t)dt = R∞ 0

β^α

Γ(α)t^α−1e^−βtdt = _Γ(α)^βα R∞

0 t^α−1e^−βtdt = _Γ(α)^βα (Γ(α)β^−α) = 1 所以_fX(t)是機率密度函數 ₍習題_2.10)

3.2 因為f_X(t) = ^√1

2πσe^{−(t−µ)22σ2} ≥ 0, t ∈ R, 而且 R∞

−∞f_X(t)dt = R∞

−∞

√1

2πσe^{−(t−µ)22σ2} dt = ^√_2πσ¹ R∞

−∞e^{−(t−µ)22σ2} dt = ^√1_π R∞

−∞e^−u2du = ^√1_π√ π

= 1 (u = ^√t−µ_2σ)

所以_fX(t)是機率密度函數

1

3.3 E(X) =R∞

−∞t^√1

2πσe^{−(t−µ)22σ2} dt = ^√1

2πσ

R∞

−∞te^{−(t−µ)22σ2} dt = ^√1

2πσ

√2πσµ = µ (習題_2.9)l

E(X²) =R∞

−∞t^2√1

2πσe^{−(t−µ)22σ2} dt = ^√1

2πσ

R∞

−∞t²e^{−(t−µ)22σ2} dt = ^√1

2πσ

√2πσ(µ²+ σ²)

= µ²+ σ²

所以V ar(X) = E(X²) − E²(X) = µ²+ σ²− µ² = σ²

3.4 E(Y ) = E(^X−µ_σ ) = _σ¹E(x − µ) = _σ¹(E(X) − µ) = 0 V ar(Y ) = V ar(^X−µ_σ ) = _σ¹2V ar(X − µ) = _σ¹2V ar(X) = 1

3.5 因為_fX(t) ≥ 0, t ≥ 0, 而且 ^R∞

0 λe^−λtdt = lim

a→∞(Ra

0 λe^−λtdt) = lim

a→∞(−e^−λt|^a₀) = 1 所以_fX(t)是機率密度函數

E(X) =R∞

0 tλe^−λtdt = lim

a→∞(Ra

0 tλe^−λtdt) = lim

a→∞[(−te^−λt)|^a₀ −Ra

0 −e^−λtdt]

= lim

a→∞[−ae^−λa+ (−¹_λe^−λt)|^a₀] = lim

a→∞(−ae^−λa−_λ¹e^−λa+¹_λ) = _λ¹ E(X²) =R∞

0 t²λe^−λtdt = lim

a→∞(Ra

0 t²λe^−λtdt) = lim

a→∞[(−t²e^−λt)|^a₀−Ra

0 −2te^−λtdt]

= 2 lim

a→∞

Ra

0 2te^−λtdt = 2 lim

a→∞[(−_λ¹te^−λt)|^a₀−Ra

0 −¹_λe^−λtdt] = 2 lim

a→∞[−_λ¹ae^−λa+(−_λ¹2e^−λt)|^a₀]

= 2 lim

a→∞(−_λ¹ae^−λa− _λ¹2e^−λa+ _λ¹2) = _λ²2

V ar(X) = E(X²) − E²(X) = _λ²2 − _λ¹2 = _λ¹2

3.7 (1)

F_X+α(t) = P (X + α ≤ t) = P (X ≤ t − α) = Rt−α

−∞ f_X(s)ds 所以_{fX+α(t) = FX+α}⁰ _{(t) = fX}(t − α) · (t − α)⁰ = fX(t − α), t ∈ R (2)

F_αX(t) = P (αX ≤ t) = P (X ≤ _α^t) =R _α^t

−∞f_X(s)ds 所以_{fαX(t) = FαX}⁰ _{(t) = fX(}^t

α) · (_α^t)⁰ = _α¹f_X(_α^t), t ∈ R

3.8 (1)

由性質_3.2知 f_Z(t) =R∞

−∞f_X(t−s)f_Y(s)ds =R∞

−∞

√1

2πσe^{−(t−s−µ)22σ2} ·^√_2πσ¹ e^{−(s−µ)22σ2} ds = _2πσ¹2

R∞

−∞e⁻⁽(t−s−µ)2+(s−µ)2

2σ2 )ds

= _2πσ¹2e^{−(t−2µ)24σ2} R∞

−∞e⁻

(s− 12t)2

σ2 ds = _2πσ¹ e^{−(t−2µ)24σ2} R∞

−∞e^−u2du = _2πσ¹ e^{−(t−2µ)24σ2} √

π = ^√1

4πσe^{−(t−2µ)24σ2}

2

(2)

令_{Zn = X1+ X2} + · · · + X_n 由f_Z1(t) = ^√1

2πσe^{−(t−µ)22σ2} , f_Z2(t) = ^√1

4πσe^{−(t−2µ)24σ2} 可猜測_{fZn =} ^{√ 1}

2nπσe^{−(t−nµ)22nσ2} , 以下將用數學歸納法證明 (I)n = 1, f_Z1(t) = ^√_2πσ¹ e^{−(t−µ)22σ2}

(II) 假設 _{n = k}時_,fZ

k(t) = ^{√ 1}

2kπσe^{−(t−kµ)22kσ2}

因為_Xi為獨立隨機變數_,所以_{Zk, Xk+1}亦為獨立_, 因此 f_Zk+1(t) = f_Zk+Xk+1(t) =R∞

−∞f_Zk(t − s)f_Xk+1(s)ds

=R∞

−∞

√ 1

2kπσe^{−(t−s−kµ)22kσ2} · ^√1

2πσe^{−(s−µ)22σ2} ds = ^{√ 1}

4kπ²σ²

R∞

−∞e^{−((t−s−kµ)22kσ2 +(s−µ)22σ2 )}ds

= ^{√ 1}

4kπ²σ²e⁻

(t−(k+1)µ)2 2(k+1)σ2 R∞

−∞e

−^(s−

t k+1)2 2k k+1σ2

ds = ^{√ 1}

4kπ²σ²e⁻

(t−(k+1)µ)2 2(k+1)σ2 R∞

−∞

q 2k

k+1σe^−u2du

= ^{√ 1}

4kπ²σ²e⁻

(t−(k+1)µ)2 2(k+1)σ2

q 2k k+1σ√

π = √ ¹

2(k+1)πσe⁻

(t−(k+1)µ)2 2(k+1)σ2

所以由數學歸納法可知_,fZ

n(t) = ^√1

2nπσe^{−(t−nµ)22nσ2}

3.9 當t < 0, F_Y(t) = P (Y ≤ t) = P (X² ≤ t) = 0 因此_{fY(t) = FY}⁰ (t) = 0, t < 0.

當t > 0, FY(t) = P (Y ≤ t) = P (X² ≤ t) = P (−t¹² ≤ X ≤ t¹²) = Rt¹²

−t¹²

√1

2πe^−s22 ds 因此f_Y(t) = F_Y⁰ (t) = ^√1

2πe^−2t(t¹²)⁰− ^√1

2πe^−2t(−t¹²)⁰ = ¹

2√

2πe^−2tt⁻¹² + ¹

2√

2πe^−2tt⁻¹²

= ^√1

2πe^−2tt⁻¹², t > 0.

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